# Why is the heaviside function in the inverse Laplace transform of 1?

• Blanchdog
In summary, the Laplace transform of a function f(t), defined on the real numbers, is ##\int_0^\infty f(t)e^{-st}dt##. The Heaviside function, H(x), is defined to be 0 for x < 0, 1/2 for x = 0, and 1 for x > 0. For the interval in the Laplace integral, the Heaviside function is identical to the constant function f(x) = 1. However, in the context of a larger problem involving the inverse Laplace transform of 1/s, the Heaviside function is multiplied by the inverse transform of 1/s with t substituted by (t-c), in this
Blanchdog
Homework Statement:: Why is the heaviside function in the inverse laplace transform of 1?
Relevant Equations:: N/A

This is a small segment of a larger problem I've been working on, and in my book it gives the transform of 1 as 1/s and vice versa. But as I've looked online for help in figuring parts of this out, I keep seeing that the heaviside function is the inverse laplace transform of 1/s, when I would think (according to my book) that it should just be 1. Can anyone explain this?

Blanchdog said:
Homework Statement:: Why is the heaviside function in the inverse laplace transform of 1?
Relevant Equations:: N/A

This is a small segment of a larger problem I've been working on, and in my book it gives the transform of 1 as 1/s and vice versa. But as I've looked online for help in figuring parts of this out, I keep seeing that the heaviside function is the inverse laplace transform of 1/s, when I would think (according to my book) that it should just be 1. Can anyone explain this?
The Laplace transform of a function f(t), defined on the real numbers, is ##\int_0^\infty f(t)e^{-st}dt##. The Heaviside function, H(x), is defined to be 0 for x < 0, 1/2 for x = 0, and 1 for x > 0. For the interval in the Laplace integral, the Heaviside function is identical to the constant function f(x) = 1.

Mark44 said:
The Laplace transform of a function f(t), defined on the real numbers, is ##\int_0^\infty f(t)e^{-st}dt##. The Heaviside function, H(x), is defined to be 0 for x < 0, 1/2 for x = 0, and 1 for x > 0. For the interval in the Laplace integral, the Heaviside function is identical to the constant function f(x) = 1.

Okay, so in the context of the broader problem: $$x=L^{-1} \frac{e^(-2s)}{s(s+2)^2}$$ In which I get the result $$x = \frac{H(t-2)}{4}(1 -e^{4-2t}+2te^{4-2t}-4e^{4-2t})$$ The first term could be replaced by ##H(t-2)## and both would be correct?

Blanchdog said:
Okay, so in the context of the broader problem: $$x=L^{-1} \frac{e^(-2s)}{s(s+2)^2}$$ In which I get the result $$x = \frac{H(t-2)}{4}(1 -e^{4-2t}+2te^{4-2t}-4e^{4-2t})$$ The first term could be replaced by ##H(t-2)## and both would be correct?
I'm not sure what you're doing above.
$$\mathcal L^{-1}[\frac{e^{-2s}}{s(s + 2)^2}] = \mathcal L^{-1}[ \frac{1/4e^{-2s}}{s} + \frac{-1/4e^{-2s}}{s+2} + \frac{-1/2e^{-2s}}{(s + 2)^2}]$$
I don't know why you have four terms inside the parentheses or why you have H(t-2) multiplying those four terms.

Mark44 said:
I'm not sure what you're doing above.
$$\mathcal L^{-1}[\frac{e^{-2s}}{s(s + 2)^2}] = \mathcal L^{-1}[ \frac{1/4e^{-2s}}{s} + \frac{-1/4e^{-2s}}{s+2} + \frac{-1/2e^{-2s}}{(s + 2)^2}]$$
I don't know why you have four terms inside the parentheses or why you have H(t-2) multiplying those four terms.
I broke up the last term because it was multiplied by ##(t-1)##. I also got the last term as being positive rather than negative in my partial fraction expansion.

EDIT: Found my algebra error, your partial fraction expansion is right.

Last edited:
I didn't work out the inverse Laplace transforms of what I showed. All I did was to get the partial fraction decomposition.
$$\frac 1 {s(s + 2)^2} = \frac {1/4}{s} + \frac{-1/4}{s+ 2} + \frac{-1/2}{(s + 2)^2}$$
I've checked my work, so I'm pretty confident of the coefficients.

I don't understand what you're saying about being multiplied by t - 1. I also don't understand why you have H(t - 2) multiplying all of your terms.

I've used the theorem ##L (H(t-c)*f(t-c)) = e^{-cs} F(s)##, where the partial fraction expansion above is ##F(s)##, and ##c = 2##. Since the problem is of the form of this theorem, the inverse transform results in a heaviside function multiplied by the inverse transform of ##F(s)## with t substituted by ##(t-c)##, in this case ##(t-2)## (not 1, I misread my work before).

Blanchdog said:
I've used the theorem L(H(t−c)∗f(t−c))=e−csF(s), where the partial fraction expansion above is F(s), and c=2. Since the problem is of the form of this theorem, the inverse transform results in a heaviside function multiplied by the inverse transform of F(s) with t substituted by (t−c), in this case (t−2) (not 1, I misread my work before).
Be careful when you write ##\mathcal L[H(t - c) * f(t - c)]##. I took this at first to mean the transform of the convolution of H(t - c) and f(t - c). Not that the symbol '*' is used for convolution.
Some definitions:
$$f(t) * g(t) = \int_0^1 f(t - x)g(x)dx$$
$$\mathcal L[f(t) * g(t)] = F(s)G(s)$$

Regarding your problem, with the wrinkles removed, do you still have any questions?

Last edited:
When you take a Laplace Transform of a function ##f##, you typically ignore what ##f(t)## is for negative values of ##t##, so when the table says ##f(t)=1##, what it really means is ##f(t)=1## for ##t\ge 0##. But when you shift this function to the right, the value of ##f(t)## for negative values of ##t## becomes important, and it's assumed that ##f(t)=0## for ##t<0##. So the full definition of ##f## is really
$$f(t) = \begin{cases} 1 & t \ge 0 \\ 0 & t<0 \end{cases},$$ which you should recognize is one of the possible definitions for the Heaviside step function.

On a related note, when you invert, say, ##F(s) = \frac 1{s+a}## and get ##f(t) = e^{-at}##, it's understood that this result is valid only for ##t\ge 0##. Technically, you should write ##f(t) = e^{-at} H(t)##, but pedantically including the step function can get tedious, so it's often omitted and assumed that the reader knows it's there.

## 1. What is the Heaviside function and how is it related to the inverse Laplace transform of 1?

The Heaviside function, also known as the unit step function, is a mathematical function that is defined as 0 for negative inputs and 1 for positive inputs. It is commonly denoted by the symbol H(t). In the context of Laplace transforms, the Heaviside function is used to represent a sudden change or "step" in a system. In the inverse Laplace transform of 1, the Heaviside function is used to indicate that the system has a constant input of 1, representing a step input function.

## 2. Why is the Heaviside function used in the inverse Laplace transform of 1 instead of other functions?

The Heaviside function is used in the inverse Laplace transform of 1 because it is the simplest way to represent a step function. Other functions, such as the rectangular function or the ramp function, can also be used to represent a step function, but they are more complex and not as commonly used in mathematical equations.

## 3. How does the Heaviside function affect the shape of the inverse Laplace transform of 1?

The Heaviside function does not directly affect the shape of the inverse Laplace transform of 1. Instead, it is used to indicate the point at which the step function occurs. For example, if the inverse Laplace transform of 1 is represented by F(t), the Heaviside function would be multiplied by F(t) to indicate when the step function occurs.

## 4. Can the Heaviside function be used in other types of mathematical equations?

Yes, the Heaviside function can be used in a variety of mathematical equations, not just in the context of Laplace transforms. It is commonly used in differential equations, signal processing, and control systems to represent a sudden change or step in a system.

## 5. Are there any limitations or drawbacks to using the Heaviside function in the inverse Laplace transform of 1?

One potential limitation of using the Heaviside function in the inverse Laplace transform of 1 is that it assumes a perfect step function, which may not always be the case in real-world systems. Additionally, the Heaviside function can sometimes lead to discontinuities in the inverse Laplace transform, which can be mathematically challenging to work with. However, these limitations can often be mitigated by using alternative functions or techniques in the inverse Laplace transform.

Replies
5
Views
1K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
1
Views
2K
Replies
10
Views
2K
Replies
4
Views
2K
Replies
7
Views
3K
Replies
9
Views
2K